\newproblem{lay:5_3_31}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.3.31}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Construct a $2\times 2$ matrix that is invertible but not diagonalizable.
}{
   % Solution
	The matrix $A$ below is such a matrix
	\begin{center}
		$A=\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$
	\end{center}
	Its inverse is
	\begin{center}
		$A^{-1}=\begin{pmatrix}1 & -1 \\ 0 & 1\end{pmatrix}$
	\end{center}
	But it is not diagonalizable. Let's see why. Let's calculate the eigenspace associated to $\lambda=1$.
	\begin{center}
		$(A-I)\mathbf{v}=\mathbf{0}$ \\
		$\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\mathbf{v}=\mathbf{0}$
	\end{center}
	whose set of solutions is formed by all vectors of the form $\mathbf{v}=(x_1,0)$ and its basis is $\{(1,0)\}$. Since the dimension of the eigenspace is 1 and there are
	2 columns in $A$, by Theorem 6.3.7, the matrix is not diagonalizable.
}
\useproblem{lay:5_3_31}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
